∫ 1 Cos ( X + a ) Cos ( X + B ) D X - Mathematics

Sum
$\int\frac{1}{\cos\left( x + a \right) \cos\left( x + b \right)}dx$

Solution

$\int\frac{1}{\cos\left( x + a \right) \cos\left( x + b \right)}dx$
$\text{Multiplying and Dividing by} \sin\left[ \left( x + b \right) - \left( x + a \right) \right], \text{we get}$
$= \int\frac{1}{\sin\left[ \left( x + b \right) - \left( x + a \right) \right]} \times \frac{\sin\left[ \left( x + b \right) - \left( x + a \right) \right]}{\cos\left( x + a \right) \cos\left( x + b \right)}dx$
$= \int\frac{1}{\sin\left( b - a \right)} \times \frac{\sin\left[ \left( x + b \right) - \left( x + a \right) \right]}{\cos\left( x + a \right) \cos\left( x + b \right)}dx$

$= \frac{1}{\sin\left( b - a \right)}\int\frac{\sin\left( x + b \right)\cos\left( x + a \right) - \sin\left( x + a \right)\cos\left( x + b \right)}{\cos\left( x + a \right) \cos\left( x + b \right)}dx$
$= \frac{1}{\sin\left( b - a \right)}\left[ \int\frac{\sin\left( x + b \right)}{\cos\left( x + b \right)}dx - \int\frac{\sin\left( x + a \right)}{\cos\left( x + a \right)}dx \right]$
$= \frac{1}{\sin\left( b - a \right)}\left[ \int\tan\left( x + b \right)dx - \int\tan\left( x + a \right)dx \right]$
$= \frac{1}{\sin\left( b - a \right)}\left[ \log\left( \sec\left( x + b \right) \right) - \log\left( \sec\left( x + a \right) \right) \right] + c$
$= \frac{1}{\sin\left( b - a \right)}\left[ \log\left( \frac{\sec\left( x + b \right)}{\sec\left( x + a \right)} \right) \right] + c$

Hence , $\int\frac{1}{\cos\left( x + a \right) \cos\left( x + b \right)}dx = \frac{1}{\sin\left( b - a \right)}\left[ \log\left( \frac{\sec\left( x + b \right)}{\sec\left( x + a \right)} \right) \right] + c$

Concept: Evaluation of Simple Integrals of the Following Types and Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.8 | Q 28 | Page 48